A) Velocity at \[t=8s\]is\[27\,m/s\]
B) Velocity is maximum at \[t=8s\]
C) Velocity at t = 12 is 20 m/s
D) All of the above.
Correct Answer: D
Solution :
Let the velocity at the end of 4 s be\[{{u}_{1}}.\] Using \[v=u+\] at from t = 0 to t = 4 s yields \[{{u}_{1}}=5\times 4=20\,m/s\] Let the velocity at the end of 8 s be \[{{u}_{2}}.\]Invoking \[v=u+at\]at from \[t=4\]to \[t=8\,s\]yields. \[{{u}_{2}}={{u}_{1}}+2\times 4\Rightarrow {{u}_{2}}=20+2\times 4=28\,m/s.\] The acceleration remains positive from t = 0 to \[t=8\,s.\]Hence the magnitude of the velocity is maximum at\[t=8\,s.\] Let the velocity of the body be \[{{u}_{3}}\] at the end of 12 s. Invoking \[v=u+\]at from \[t=8\,s\]to \[t=12\,s\]yields \[{{u}_{3}}=28-2\times 4=20\,m/s.\] Hence, the correction option is (d).
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